Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes.

Can you answer this question?

1 Answer

0 votes
  • To check if a given distribution is a probability distribution of random variable, the sum of the individual probabilties should add up to 1 (i.e, $\sum P(X_i) = 1$). Also 0 $\lt$ P(X) $\leq$ 1.
Step 1:
Let $X$ denote the number of doublets.
Possible doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
Clearly $X$ can take values as $0,1,2,3$ and 4
Probability of getting a doublet =$\large\frac{6}{36}=\frac{1}{6}$
Probability of not getting a doublet =$1-\large\frac{1}{36}=\frac{1}{6}$
Step 2:
Now $P(X=0)=P$(no doublets)
$\qquad\qquad\quad\;\;=\large\frac{5}{6}\times \frac{5}{6}\times\frac{5}{6}\times \frac{5}{6}$
Step 3:
$P(X=1)=P$(one doublet and three non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6}\times \frac{5^3}{6^3}\big)$
Step 4:
$P(X=2)=P$(two doublet and two non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^2}\times \frac{5^2}{6^2}\big)$
Step 5:
$P(X=3)=P$(three doublets and one non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^3}\times \frac{5}{6}\big)$
Step 6:
$P(X=4)=\large\frac{1}{6}\times \large\frac{1}{6}\times\large\frac{1}{6}\times\large\frac{1}{6}$
Thus the required probability distribution is
answered Sep 27, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App